#15 Deriving Poisson's and Laplace's equations from Gauss's law, the Laplacian in three coordinate systems, a boundary-value workflow, and an electrohydrodynamic pump example

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Move from dielectrics to potential equations

   

   In our last lesson, we explored polarization and dielectrics.

   We defined the polarization vector P, derived bound charge densities, and introduced the dielectric constant ε_r and permittivity ε.

   We also established boundary conditions for D and E at dielectric interfaces.

   Today, we move to one of the most powerful tools in electrostatics: Poisson's and Laplace's equations.

   These equations let us find the electric potential V when we know only the boundary conditions, not the charge distribution everywhere.

   Narration transcript

   In our last lesson, we explored polarization and dielectrics. We defined the polarization vector P, derived bound charge densities, and introduced the dielectric constant epsilon r and permittivity epsilon. We also established boundary conditions for D and E at dielectric interfaces. Today, we move to one of the most powerful tools in electrostatics: Poisson's and Laplace's equations. These equations let us find the electric potential V when we know only the boundary conditions, not the charge distribution everywhere.

2. 2. Derive Poisson's and Laplace's equations

   

   Let's derive these equations from what we already know.

   Start with Gauss's law:

   $\displaystyle ∇\cdot D = \rho _{v}.$

   For a linear, isotropic, homogeneous medium D = εE, so ∇·(εE) = ρ_v.

   We also know E = −∇V.

   Substitution gives ∇·(−ε∇V) = ρ_v.

   If ε is constant throughout the region, we can pull it out: negative ε times the ∇²V equals ρ_v.

   Poisson's equation:

   $\displaystyle ∇²V = -\rho _{v}/\varepsilon .$

   When there is no free charge in the region, ρ_v equals zero, and Poisson's equation simplifies to Laplace's equation: ∇²V equals zero.

   Narration transcript

   Let's derive these equations from what we already know. We start with Gauss's law in differential form: divergence of D equals the volume charge density rho v. For a linear, isotropic, homogeneous medium, D equals epsilon E, so divergence of epsilon E equals rho v. We also know that E equals negative gradient of V. Substituting this into Gauss's law, we get: divergence of negative epsilon gradient V equals rho v. If epsilon is constant throughout the region, we can pull it out: negative epsilon times the Laplacian of V equals rho v. This gives us Poisson's equation: Laplacian of V equals negative rho v over epsilon. When there is no free charge in the region, rho v equals zero, and Poisson's equation simplifies to Laplace's equation: Laplacian of V equals zero.

3. 3. Write the Laplacian in three coordinate systems

   

   The ∇²V takes different forms in each coordinate system.

   In Cartesian coordinates, it is the sum of the second partial derivatives with respect to x, y, and z.

   In cylindrical coordinates, the Laplacian has a one over rho term with a rho derivative, a one over rho squared term with the phi derivative, and a z derivative.

   In spherical coordinates, it has a one over r squared term with an r derivative, a one over r squared sine theta term with the theta derivative, and a one over r squared sine squared theta term with the phi derivative.

   The general procedure to solve a boundary-value problem is: Step one, solve Laplace's or Poisson's equation by direct integration to get V with unknown constants.

   Step two, apply boundary conditions to determine those constants.

   Step three: find E from E = −∇V.

   Step four, if needed, find D, surface charge, or capacitance.

   Narration transcript

   The Laplacian of V takes different forms in each coordinate system. In Cartesian coordinates, it is the sum of the second partial derivatives with respect to x, y, and z. In cylindrical coordinates, the Laplacian has a one over rho term with a rho derivative, a one over rho squared term with the phi derivative, and a z derivative. In spherical coordinates, it has a one over r squared term with an r derivative, a one over r squared sine theta term with the theta derivative, and a one over r squared sine squared theta term with the phi derivative. The general procedure to solve a boundary-value problem is: Step one, solve Laplace's or Poisson's equation by direct integration to get V with unknown constants. Step two, apply boundary conditions to determine those constants. Step three, find E from E equals negative gradient of V. Step four, if needed, find D, surface charge, or capacitance.

4. 4. Solve the EHD pump example

   

   Let's work through Example 6.1 from Sadiku.

   An electrohydrodynamic pump has two parallel electrodes separated by distance d.

   The region between them contains a uniform charge density ρ₀.

   The left electrode is at voltage V₀, and the right electrode at zero volts.

   Find the pressure of the pump.

   Given ρ₀ = 25 mC/m³.

   The left electrode is at V₀ = 22 kV.

   Step one: Since ρ_v is not zero, we use Poisson's equation.

   The potential depends only on z, so the Laplacian reduces to d² V over d z² equals negative ρ₀ over ε.

   Step two: Integrate once.

   d V over d z equals negative ρ₀ z over ε, plus a constant A.

   Step three: Integrate again.

   V equals negative ρ₀ z² over 2 ε, plus A z, plus B.

   Step four: Apply boundary conditions.

   V(0) = V₀, so B = V₀.

   Thus B = 22,000 V.

   V(d) = 0 ⇒ 0 = −ρ₀d²/(2ε) + Ad + V₀.

   $\displaystyle A =\frac{ \rho ₀d}{2\varepsilon }- V₀/d.$

   Step five: E = −dV/dz.

   $\displaystyle E = \left(\frac{\rho ₀z}{\varepsilon }- A\right)a_{z}.$

   The net force is F = ∫ρ₀E dv.

   The integral gives F = ρ₀SV₀a_z.

   Pressure is p = F/S = ρ₀V₀.

   p = 25×10⁻³×22×10³ Pa.

   The result is p = 550 N/m² = 550 Pa.

   Narration transcript

   Let's work through Example 6.1 from Sadiku. An electrohydrodynamic pump has two parallel electrodes separated by distance d. The region between them contains a uniform charge density rho o. The left electrode is at voltage V o, and the right electrode at zero volts. Find the pressure of the pump. Given values: rho o equals 25 milliCoulombs per cubic metre. V o equals 22 kilovolts. Step one: Since rho v is not zero, we use Poisson's equation. The potential depends only on z, so the Laplacian reduces to d squared V over d z squared equals negative rho o over epsilon. Step two: Integrate once. d V over d z equals negative rho o z over epsilon, plus a constant A. Step three: Integrate again. V equals negative rho o z squared over 2 epsilon, plus A z, plus B. Step four: Apply boundary conditions. At z equals 0, V equals V o, so B equals V o. That gives us B equals 22000 volts. At z equals d, V equals 0, so zero equals negative rho o d squared over 2 epsilon, plus A d, plus V o. Solving for A: A equals rho o d over 2 epsilon, minus V o over d. Step five: The electric field is E equals negative d V over d z. This gives E equals rho o z over epsilon minus A, in the z direction. The net force on the charge is F equals the integral of rho o times E over the volume. After evaluating, the force simplifies to F equals rho o times S times V o, in the z direction. The pressure is force per unit area: p equals rho o times V o. Substituting: p equals 25 times 10 to the negative 3, times 22 times 10 to the 3. This gives p equals 550 Newtons per square metre.

5. 5. Review the Poisson–Laplace workflow

   

   Let's summarize.

   Poisson's equation, ∇²V equals negative ρ_v over ε, applies when free charge is present.

   Laplace's equation ∇²V = 0 applies in charge-free regions.

   The solution procedure is: integrate the equation, apply boundary conditions, then find E from the gradient of V.

   In our next lesson, we'll apply these equations to find resistance and capacitance of various geometries.

   Narration transcript

   Let's summarize. Poisson's equation, Laplacian of V equals negative rho v over epsilon, applies when free charge is present. Laplace's equation, Laplacian of V equals zero, applies in charge-free regions. The solution procedure is: integrate the equation, apply boundary conditions, then find E from the gradient of V. In our next lesson, we'll apply these equations to find resistance and capacitance of various geometries.

Source video: Electromagnetic Theory (v2) #15 Poisson's & Laplace's Equations (5:49)